The integrand blows up near x = 1 {\displaystyle x=1} , so this is an improper integral:
∫ 0 1 1 ( 1 − x ) 3 / 2 d x = lim t → 1 − ∫ 0 t ( 1 − x ) − 3 / 2 d x . {\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {1}{(1-x)^{3/2}}}\;\mathrm {d} x&=\lim _{t\to 1^{-}}\int _{0}^{t}(1-x)^{-3/2}\;\mathrm {d} x.\end{aligned}}}
To find the indefinite integral, use the u-substitution u = 1 − x {\displaystyle u=1-x} , d u = − d x {\displaystyle \mathrm {d} u=-\mathrm {d} x} :
∫ 1 ( 1 − x ) 3 / 2 d x = − ∫ u − 3 / 2 d u = − ( − 2 ) u − 1 / 2 + c = 2 ( 1 − x ) − 1 / 2 + c = 2 1 − x + c . {\displaystyle {\begin{aligned}\int {\frac {1}{(1-x)^{3/2}}}\;\mathrm {d} x&=-\int u^{-3/2}\mathrm {d} u=-(-2)u^{-1/2}+c=2(1-x)^{-1/2}+c={\frac {2}{\sqrt {1-x}}}+c.\end{aligned}}}
Thus,
∫ 0 1 1 ( 1 − x ) 3 / 2 d x = lim t → 1 − [ 2 1 − x ] | x = 0 x = t = lim t → 1 − ( 2 1 − t − 2 ) = + ∞ . {\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {1}{(1-x)^{3/2}}}\;\mathrm {d} x&=\lim _{t\to 1^{-}}\left.\left[{\frac {2}{\sqrt {1-x}}}\right]\right|_{x=0}^{x=t}=\lim _{t\to 1^{-}}\left({\frac {2}{\sqrt {1-t}}}-2\right)=+\infty .\end{aligned}}}
The integral diverges.
, so this is an improper integral:
, 
:
![{\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {1}{(1-x)^{3/2}}}\;\mathrm {d} x&=\lim _{t\to 1^{-}}\left.\left[{\frac {2}{\sqrt {1-x}}}\right]\right|_{x=0}^{x=t}=\lim _{t\to 1^{-}}\left({\frac {2}{\sqrt {1-t}}}-2\right)=+\infty .\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/273634b9cd44880ff6f4e48b34bead137205c4f8)
